\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+18\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)-16\)

\(=\left(x^2+10x+20\right)^2-16+16=\left(x^2+10x+20\right)^2\)

Chúc bạn học tốt.

      \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(\Rightarrow\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+6\right)\left(x+8\right)\right]+16\)

\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(\Rightarrow\left(x^2+10x+16\right)\left[\left(x^2+10x+16\right)+8\right]+16\)

\(\Rightarrow\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+4^2\)

\(\Rightarrow\left(x^2+10x+20\right)^2\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10+16+8\right)+16\)

\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)

\(=\left(x^2+10x+16+4\right)^2\)

\(=\left(x^2+10+20\right)^2\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right) \left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\)
\(\left(1\right)\Rightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\)
\(=\left(x^2+10x+20\right)^2\)
 

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\)

\(=\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+16\)

\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\)(1) 

Đặt \(x^2-10x+20=t\)thay vào (1) ta được : 

\(\left(t-4\right)\left(t+4\right)+16\)

\(=t^2-16+16\)

\(=t^2\)Thay \(t=x^2-10x+20\)ta được :

\(\left(x^2-10x+20\right)^2\)

\(=\left(x^2-2.5.x+25-25+20\right)^2\)

\(=\left[\left(x-5\right)^2-5\right]^2\)

\(=\left(x-5-\sqrt{5}\right)^2\left(x-5+\sqrt{5}\right)^2\)

\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

Sao đề là phân tích mà lại "= 0" vậy bạn?

\(S=x^6-8\)

\(S=\left(x^2\right)^3-2^3\)

\(S=\left(x^2-2\right)\left(x^4+2x^2+4\right)\)

⇒ Chọn C

\(=\left(x^2\right)^3-2^3=\left(x^2-2\right)\left(x^4+2x^2+4\right)\\ =>C\)

\(5x(2x+3)+6x+9\\=5x(2x+3)+3(2x+3)\\=(2x+3)(5x+3)\)

a: \(5x\left(2x+3\right)+6x+9\)

\(=5x\left(2x+3\right)+\left(6x+9\right)\)

\(=5x\left(2x+3\right)+3\left(2x+3\right)\)

\(=\left(2x+3\right)\left(5x+3\right)\)

b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+4\right)\left(3x+48+5\right)\)

=(x+4)(3x+53)

\(A=x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)

\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

Đặt \(x^2+10x=a\) nên : 

 \(A=a\left(a+24\right)+128=a^2+24a+144-16=\left(a+12\right)^2-4^2=\left(a+16\right)\left(a+8\right)\)

\(=\left(x^2+10x+8\right)\left(x^2+10x+16\right)=\left(x^2+10x+8\right)\left(x+2\right)\left(x+8\right)\)

\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)

\(=\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)

\(=\left(x-1\right)\left(x+1\right)\left(x+4+1\right)\left(x+4-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+5\right)\left(x-3\right)\)

=.= hok tốt!!